The curve y = |x|/(sqrt(5- x^2)) is called a bullet-nose curve. Find an equation of the tangent line to this curve at the point (2, 2)
Well, if this curve is called a bullet-nose curve, does that mean it's sharp and pointy like the nose of a bullet? I sure hope not! Otherwise, we might need to be careful around it.
Now, to find the equation of the tangent line to the curve at the point (2, 2), we need to determine the slope of the tangent line. We can do that by finding the derivative of the curve equation.
The derivative of |x| is just sign(x), and the derivative of (sqrt(5 - x^2)) is (-x)/sqrt(5 - x^2). So, the derivative of the curve equation with respect to x is:
dy/dx = [sign(x) * sqrt(5 - x^2) - x^2/ sqrt(5 - x^2)] / sqrt(5 - x^2)
Now, to find the slope of the tangent line at the point (2, 2), we substitute x = 2 into the derivative:
dy/dx = [sign(2) * sqrt(5 - 2^2) - 2^2 / sqrt(5 - 2^2)] / sqrt(5 - 2^2)
= [1 * sqrt(5 - 4) - 4 / sqrt(5 - 4)] / sqrt(5 - 4)
= [sqrt(1) - 4 / 1] / 1
= [1 - 4] / 1
= -3
So, the slope of the tangent line at (2, 2) is -3. Now we can use the point-slope form of a line to find the equation of the tangent line. Since the point is (2, 2) and the slope is -3, the equation is:
y - 2 = -3(x - 2)
Oh, calculating tangent lines can be as fun as a clown on a unicycle! Now we have the equation of the tangent line to the bullet-nose curve at the point (2, 2).
To find the equation of the tangent line to the curve at the point (2, 2), we need to determine the slope of the tangent line at that point. We can find the slope using the derivative of the curve.
First, let's find the derivative of the curve y = |x|/(sqrt(5 - x^2)) with respect to x.
Using the quotient rule, the derivative dy/dx is given by:
dy/dx = (d/dx (|x|) * (sqrt(5 - x^2)) - |x| * (d/dx (sqrt(5 - x^2)))) / (sqrt(5 - x^2))^2
The derivative of |x| can be written as:
d/dx (|x|) = x/|x| = x/sqrt(x^2) = x/√(x^2) = x/|x| = x/x = 1 (for x ≠ 0)
Now, let's find the derivative of sqrt(5 - x^2):
d/dx (sqrt(5 - x^2)) = (-x) / (2 * sqrt(5 - x^2))
Substituting these derivatives back into the derivative dy/dx equation, we get:
dy/dx = (1 * sqrt(5 - x^2) - |x| * (-x) / (2 * sqrt(5 - x^2))) / (sqrt(5 - x^2))^2
Simplifying further, we have:
dy/dx = (sqrt(5 - x^2) + x^2) / (2 * sqrt(5 - x^2) * (5 - x^2))
Now, we can find the slope of the tangent line by substituting x = 2 into the derivative equation:
dy/dx = (sqrt(5 - 2^2) + 2^2) / (2 * sqrt(5 - 2^2) * (5 - 2^2))
= (sqrt(5 - 4) + 4) / (2 * sqrt(5 - 4) * (5 - 4))
= (sqrt(1) + 4) / (2 * sqrt(1) * 1)
= (1 + 4) / (2 * 1 * 1)
= 5 / 2
Therefore, the slope of the tangent line at the point (2, 2) is 5/2.
Next, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope.
Substituting the values x1 = 2, y1 = 2, and m = 5/2 into the point-slope form, we get:
y - 2 = (5/2)(x - 2)
Simplifying further, we have:
y - 2 = (5/2)x - 5
Moving all terms to one side, we get:
(5/2)x - y + 3 = 0
Hence, the equation of the tangent line to the curve at the point (2, 2) is (5/2)x - y + 3 = 0.
To find the equation of the tangent line to the curve, we need to find the slope of the tangent line at the point (2, 2) and then use the point-slope form of a line to determine the equation.
1. Start by finding the derivative of the given curve, y = |x|/(sqrt(5 - x^2)). To simplify the calculation, we can rewrite the absolute value function as a piecewise function:
y = x/(sqrt(5 - x^2)) if x >= 0
y = -x/(sqrt(5 - x^2)) if x < 0
2. Taking the derivative separately for each piece, let's differentiate y = x/(sqrt(5 - x^2)) and y = -x/(sqrt(5 - x^2)).
For y = x/(sqrt(5 - x^2)), apply the quotient rule. Let u = x and v = sqrt(5 - x^2):
dy/dx = (v * du/dx - u * dv/dx) / v^2
dv/dx = d/dx (sqrt(5 - x^2)) = (1/2) * (5 - x^2)^(-1/2) * (-2x) = -x/(sqrt(5 - x^2))
du/dx = 1
Substituting these values back into the quotient rule equation, we get:
dy/dx = (sqrt(5 - x^2) - x * (-x/(sqrt(5 - x^2)))) / (sqrt(5 - x^2))^2
Simplifying further, we have:
dy/dx = (sqrt(5 - x^2) + x^2) / (5 - x^2)
Similarly, applying the same steps for y = -x/(sqrt(5 - x^2)), we get dy/dx = (sqrt(5 - x^2) - x^2) / (5 - x^2).
3. Now that we have the derivative dy/dx, we can find the slope of the tangent line at the point (2, 2) by substituting x = 2 into the derivative equation:
m = dy/dx (at x = 2) = (sqrt(5 - 2^2) + 2^2) / (5 - 2^2) = (sqrt(1) + 4) / (5 - 4) = (1 + 4) / 1 = 5
Therefore, the slope of the tangent line at (2, 2) is 5.
4. Now that we have the slope, we can use the point-slope form of a line to determine the equation of the tangent line. The point-slope form is:
y - y1 = m(x - x1)
Substituting the values m = 5, x1 = 2, and y1 = 2, we get:
y - 2 = 5(x - 2)
Simplifying and putting the equation in slope-intercept form, we have:
y = 5x - 8
So, the equation of the tangent line to the curve at the point (2, 2) is y = 5x - 8.
Here is what your curve looks like
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7Cx%7C%2F%28sqrt%285-+x%5E2%29%29
notice that -√5 < x < √5 is our domain, with x = ±√5 as asymptotes
Since (2,2) lies in quadrant I
I will use
y = x/√(5 - x^2) = x(5-x^2)^(-1/2)
dy/dx = x(-1/2)(5-x^2)^(-3/2) (-2x) + (5-x^2)^(-1/2)
at (2,2)
dy/dx = 2(-1/2)(1^(-3/2) (-4) + 1
= 4+1 = 5
so y = 5x + b
at (2,2)
2 = 10 + b
b = -8
tangent equation: y = 5x - 8
verification:
http://www.wolframalpha.com/input/?i=plot+y+%3D+%7Cx%7C%2F%28sqrt%285-+x%5E2%29%29+%2C+y+%3D+5x-8+from+1+to+2.2
well, if x>=0, |x|=x, so we have
y = x/√(5-x^2)
y' = 5/(5-x^2)^(3/2)
At x=2, y'=5
So, the line through (2,2) with slope=5 is
y-2 = 5(x-2)
See this at
http://www.wolframalpha.com/input/?i=plot+y%3D|x|%2F%E2%88%9A%285-+x^2%29%2C+y%3D5%28x-2%29%2B2+for+0%3C%3Dx%3C%3D3